JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
\(\mathrm{NO}_2\) required for a reaction is produced by decomposition of \(\mathrm{N}_2 \mathrm{O}_5\) in \(\mathrm{CCl}_4\) as by equation \(2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\)The initial concentration of \(\mathrm{N}_2 \mathrm{O}_5\) is \(3 \mathrm{~mol} \mathrm{~L}^{-1}\) and it is \(2.75 \mathrm{~mol} \mathrm{~L}^{-1}\) after 30 minutes.The rate of formation of \(\mathrm{NO}_2\) is \(\mathrm{x} \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\) \(\min ^{-1}\), value of \(x\) is _______.
- A \(16\)
- B \(17\)
- C \(18\)
- D \(19\)
Answer & Solution
Correct Answer
(B) \(17\)
Step-by-step Solution
Detailed explanation
Rate of reaction (ROR) \(=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}=\frac{1}{4} \frac{\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{O}_2\right]}{\Delta \mathrm{t}}\)…
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